We can calculate area of the regular nonagon by applying three different methods.

1)

If the length of the side of the nonagon is known.Consider one of the triangles .Draw the altitude (perpendicular ) from the apex to the base of the triangle.(The side of the nonagon forms the base.) Let the height of the altitude be h, base of the triangle be s .

The angle at the apex =

$\frac{360^o}{n}$, where n: side of the nonagon = 9

Angle at apex =

$\frac{360^o}{9}$ =400 Since t =

$\frac{1}{2}$ the angle at vertex t=200.

Consider tan ratio;

tan t =

$\frac{(s/2)}{h}$ Solving for h,

tan t =

$\frac{(s/2)}{h}$ =

$\frac{s}{2h}$h=

$\frac{s}{2}$ tan t

Area of the triangle = $\frac{1}{2}$ $\times base \times height$ =

$\frac{1}{2} \times s \times h$Consider h in terms of s,

Area =

$\frac{1}{2}$$\times s \times$ (

$\frac{s}{2tan t}$)

=

$\frac{s^2}{(4 tan t)}$ Since for the regular nonagon , t =20

^{o}, tan t =tan 20

^{o} = 0.364

So, area =

$\frac{s^2}{4}$ $\times$ 0.364 = 0.6869 $\times$ s

^{2}As there are 9 such triangles in a nonagon, the total area of the nonagon = 9 $\times$ area of the triangle.

Area of nonagon = 9 $\times$ 0.6869 $\times$ s

^{2} ** =6.1818 $\times$ s**^{2}.2)

When Radius r is known:The radius of the regular polygon is the distance between the center and the vertex of the polygon (radius of the circumcircle). In the triangle, the side of the isosceles triangle is the radius. Draw the perpendicular h (height of the triangle.) Let x be the half of the base of the triangle.

Let us write x and h in terms of r.

sin t =

$\frac{x}{r}$So , x = r sin t

cos t =

$\frac{h}{r}$h = r cos t

Area of the triangle = $\frac{1}{2}$$\times base \times height$ =x $\times$ h (since x =

$\frac{1}{2}$ base)

In terms of r,

A = r sin t $\times$ r cos t

=

$\frac{1}{2}$ r

^{2} sin(2t)

{Use double angle formula for sin : sin (2t) = 2 sin t $\times$ cos t}

Total area for the nonagon = 9 $\times$ area of the triangle

= 9 $\times$ r

^{2} $\times$

$\frac{sin (2t)}{2}$Plugging t =

$\frac{1800}{n}$ =

$\frac{1800}{9}$ = 200

2t =400

sin 40 = 0.6428

Area = 9 $\times$

$\frac{r^2}{2}$ $\times$ 0.6428

A = 2.8925 r^{2}3)

When the apothem (radius of the incircle) is given Apothem of the nonagon is the perpendicular distance from the center to the side.

Let a be the length of the apothem, s be the side.

Area of the triangle =

$\frac{!}{2}$ $\times$ a $\times$ s

Area of the triangle =

$\frac{1}{2}$$\times a \times s$

Let us show a in terms of s

tan t =

$\frac{(s/2)}{a}$so s = 2 $\times$ a $\times$ tan t

since for a regular nonagon t = 20

^{o},

so s=2a tan 20

= 2 $\times$ 0.364 a

A=

$\frac{1}{2}$ s $\times$ a

Plugging in s = 0.728 a

A=

$\frac{1}{2}$ $\times$ (0.728 a) $\times$ a

A=0.364 a

^{2}So total area of the nonagon = 9 $\times$ area of the triangle

= 9 $\times$ 0.364 (a

^{2})

=

**3.2757 a**^{2}